Question Number 175653 by mnjuly1970 last updated on 04/Sep/22
$$ \\ $$$$\:\:\:{Q}:\:\:{prove}\:{that}\:{the}\:{following} \\ $$$$\:\:\:\:\:{equation}\:{has}\:{no}\:{solution}. \\ $$$$ \\ $$$$\:\:\:\sqrt{{x}\:+\lfloor\:{x}\:\rfloor}\:+\:\sqrt{{x}\:−\sqrt{{x}}\:}\:=\:\mathrm{1} \\ $$$$ \\ $$
Answered by mahdipoor last updated on 04/Sep/22
![f(x) = (√(x−(√x) ))+ (√(x+[x])) D_f =∀x≥1 for I_n =[n,n+1) n≥1 f_n (x)=(√(x−(√x)))+(√(x+n)) D_x (f_n (x))=((1−(1/(2(√x))))/( 2(√(x−(√x)))))+(1/( 2(√(x+n)))) >0 f is incrise function in I_n ⇒ R_f in I_n =[f_n (n),f_n (n+1)) ⇒⇒ R_f in D_f =I_1 ∩I_2 ∩... = [(√2),(√3)+(√(2−(√2))))∩[(√4)+(√(2−(√2))),(√5)+(√(3−(√3))))∩... ⇒(√2)≤R_f ⇒f(x)≠1 for ∀x∈D_f](Q175659.png)
$${f}\left({x}\right)\:=\:\sqrt{{x}−\sqrt{{x}}\:}+\:\sqrt{{x}+\left[{x}\right]} \\ $$$${D}_{{f}} =\forall{x}\geqslant\mathrm{1} \\ $$$${for}\:{I}_{{n}} =\left[{n},{n}+\mathrm{1}\right)\:\:\:{n}\geqslant\mathrm{1} \\ $$$${f}_{{n}} \left({x}\right)=\sqrt{{x}−\sqrt{{x}}}+\sqrt{{x}+{n}} \\ $$$${D}_{{x}} \left({f}_{{n}} \left({x}\right)\right)=\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}}{\:\mathrm{2}\sqrt{{x}−\sqrt{{x}}}}+\frac{\mathrm{1}}{\:\mathrm{2}\sqrt{{x}+{n}}}\:>\mathrm{0} \\ $$$${f}\:{is}\:{incrise}\:{function}\:{in}\:{I}_{{n}} \Rightarrow \\ $$$${R}_{{f}} \:\:{in}\:{I}_{{n}} =\left[{f}_{{n}} \left({n}\right),{f}_{{n}} \left({n}+\mathrm{1}\right)\right) \\ $$$$\Rightarrow\Rightarrow \\ $$$${R}_{{f}} \:\:{in}\:{D}_{{f}} ={I}_{\mathrm{1}} \cap{I}_{\mathrm{2}} \cap...\:= \\ $$$$\left[\sqrt{\mathrm{2}},\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\right)\cap\left[\sqrt{\mathrm{4}}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}},\sqrt{\mathrm{5}}+\sqrt{\mathrm{3}−\sqrt{\mathrm{3}}}\right)\cap... \\ $$$$\Rightarrow\sqrt{\mathrm{2}}\leqslant{R}_{{f}} \:\:\: \\ $$$$\Rightarrow{f}\left({x}\right)\neq\mathrm{1}\:{for}\:\forall{x}\in{D}_{{f}} \\ $$
Commented by mnjuly1970 last updated on 04/Sep/22
$${thanks}\:{alot}\:{sir}\:{mahdipoor} \\ $$
Answered by mr W last updated on 04/Sep/22
$${x}\geqslant\mathrm{0} \\ $$$${x}\geqslant\sqrt{{x}}\:\Rightarrow{x}\geqslant\mathrm{1}\Rightarrow\lfloor{x}\rfloor\geqslant\mathrm{1} \\ $$$${x}+\lfloor{x}\rfloor\geqslant\mathrm{2} \\ $$$$\sqrt{{x}+\lfloor{x}\rfloor}\geqslant\sqrt{\mathrm{2}} \\ $$$$\sqrt{{x}+\lfloor{x}\rfloor}+\sqrt{{x}−\sqrt{{x}}}\geqslant\sqrt{\mathrm{2}}>\mathrm{1} \\ $$$$\sqrt{{x}+\lfloor{x}\rfloor}+\sqrt{{x}−\sqrt{{x}}}=\mathrm{1}\:{can}\:{never}\:{be}\:{true}. \\ $$$${i}.{e}.\:\sqrt{{x}+\lfloor{x}\rfloor}+\sqrt{{x}−\sqrt{{x}}}=\mathrm{1}\:{has}\:{no}\:{solution}. \\ $$
Commented by mnjuly1970 last updated on 04/Sep/22
$${bravo}\:{sir}\:{W} \\ $$